Find the sum of n terms of an A.P.

1.	Find the sum of n terms of an A.P.
Answer» Let the A.P. be a, a + d, a + 2d, ... Let l be the last term and S the required sum. Then, S = \(\frac{n}{2}\) (a + l ) = \(\frac{ Number\, of\, terms}{2}\) (First term + Last term) = \(\frac{n}{2}\) (a + a + (n – 1) d) = \(\frac{n}{2}\) (2a + (n-1)d), where n is the number of terms, a is first term and d is common difference. Also, nth term = Sum of n terms – Sum of (n – 1) terms i.e., T_n = S_n – S_n–1. Ex. The sum of 20 terms of the the A.P. 1, 3, 5, 7, 9.... is S20 = \(\frac{20}{2}\) (2 ×1 + (20 - 1)× 2) (∵ a = 1, d = 2, n = 20) = 10 × 40 = 400

Answer»

Let the A.P. be a, a + d, a + 2d, ... Let l be the last term and S the required sum.

Then,

S = \(\frac{n}{2}\) (a + l ) = \(\frac{ Number\, of\, terms}{2}\) (First term + Last term)

= \(\frac{n}{2}\) (a + a + (n – 1) d) = \(\frac{n}{2}\) (2a + (n-1)d),

where n is the number of terms, a is first term and d is common difference.

Also, nth term = Sum of n terms – Sum of (n – 1) terms

i.e., T_n = S_n – S_n–1.

Ex.

The sum of 20 terms of the the A.P. 1, 3, 5, 7, 9.... is

S20 = \(\frac{20}{2}\) (2 ×1 + (20 - 1)× 2) (∵ a = 1, d = 2, n = 20)

= 10 × 40 = 400

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